`cos^2x+cos^2(2x)+cos^2(pi/3-3x)=7/4`

Đáp án:

`S={π/6+{kπ}/3;π/{12}+{kπ}/2|k\in ZZ}`

Giải thích các bước giải:

Công thức:

`cos^2 a={1+cos2a}/2`

`cos(a-b)=cos a cosb +sin a sin b`

`sin^2 a+cos^2 a=1`

`cos a+cos b=2 cos\ {a+b}/2 cos \ {a-b}/2`

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`cos^2 x+cos^2 2x+cos^2 (π/3-3x)=7/4\ (1)`

Ta có:

`cos^2 x={1+cos 2x}/2`

`cos^2 2x={1+cos4x}/2`

`\qquad cos^2 (π/3-3x)`

`=(cos\ π/3 cos3x+sin\ π/3 sin3x)^2`

`=(1/ 2 cos 3x +\sqrt{3}/2 sin 3x)^2`

`=1/ 4 cos^2 3x +2. 1/ 2 cos3x . \sqrt{3}/2 sin 3x+3/ 4 sin^2 3x`

`=3/ 4 sin^2 3x+3/ 4 cos^2 3x-1/ 2 cos^2 3x+\sqrt{3}/2 sin 3x cos3x`

`=3/ 4 (sin^2 3x+cos^2 3x)-cos3x(cos 3x . cos\ π/ 3+sin 3x . sin\ π/3)`

`=3/ 4 . 1 -cos3x cos(3x-π/3)`

$\\$

`(1)<=>{1+ cos2x}/2 +{1+cos4x}/2+3/ 4 -cos3x cos(3x-π/3)=7/4`

`<=> 1/ 2 (cos4x +cos2x)-cos3x cos(3x-π/3)=0`

`<=>1/ 2 . 2. cos\ {4x+2x}/2 cos\ {4x-2x}/2  -cos3x cos(3x-π/3)=0`

`<=> cos3x cosx  -cos3x cos(3x-π/3)=0`

`<=> cos3x[cos x- cos(3x-π/3)]=0`

`<=>`$\left[\begin{array}{l}cos3x=0\ (2)\\cosx-cos(3x-\dfrac{π}{3})=0\ (3)\end{array}\right.$

$\\$

`(2)<=>3x=π/2+kπ`

`⇔ x=π/6+{kπ}/3\ (k\in ZZ)` 

`(3)<=>cos(3x-π/3)=cosx`

`<=>`$\left[\begin{array}{l}3x-\dfrac{π}{3}=x+k2π\\3x-\dfrac{π}{3}=-x+k2π\end{array}\right.$ `(k\in ZZ)`

`<=>`$\left[\begin{array}{l}2x=\dfrac{π}{3}+k2π\\4x=\dfrac{π}{3}+k2π\end{array}\right.$ `(k\in ZZ)`

`<=>`$\left[\begin{array}{l}x=\dfrac{π}{6}+kπ\\x=\dfrac{π}{12}+\dfrac{kπ}{2}\end{array}\right.$ `(k\in ZZ)` 

Vậy phương trình có tập nghiệm:

`S={π/6+{kπ}/3;π/{12}+{kπ}/2|k\in ZZ}`